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I came cross the following equation:
$$\lim_{x\to+\infty}\frac{1}{x}\int_0^x|\sin t|\mathrm{d}t=\frac{2}{\pi}$$ I wonder how to prove it.

Using the Mathematica I got the following result: enter image description here Could you suggest some ideas how to prove this? Any hints will be appreciated.

Angelo
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Franz Lee
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2 Answers2

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$\textbf{Hint}$: the limmand can be rewritten as

$$ \frac{\int_0^{\left\lfloor \frac{x}{\pi}\right\rfloor \pi} |\sin t|dt + \int_{\left\lfloor \frac{x}{\pi}\right\rfloor \pi}^x |\sin t|dt}{x}$$

$$=2 \frac{\left\lfloor \frac{x}{\pi}\right\rfloor}{x}+\frac{\int_{ \left\lfloor \frac{x}{\pi}\right\rfloor}^x|\sin t|dt}{x}$$

because the integrand is $\pi$-periodic. The equality $z \equiv \lfloor z \rfloor + \{z\}$ (where $0\leq \{ \cdot \} < 1$ denotes the fractional part function) can be used to calculate the limit on the left. Squeeze theorem can be used for the remainder on the right.

Ninad Munshi
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Let $x=A+\pi C$ where $A$ is between $0$ & $\pi$ , with Integer $C$.

$D(A,C) = \frac{1}{A+\pi C}\int_0^{A+\pi C}|\sin t|\mathrm{d}t$

$D(A,C) = \frac{1}{A+\pi c}\int_0^{\pi C}|\sin t|\mathrm{d}t + \frac{1}{A+\pi c}\int_{0+\pi C}^{A+\pi C}|\sin t|\mathrm{d}t$

$D(A,C) = \frac{1}{A+\pi C} [[ \int_0^{\pi C}|\sin t|\mathrm{d}t ]] + \frac{1}{A+\pi C}\int_0^{A}|\sin t|\mathrm{d}t$

$D(A,C) = \frac{1}{A+\pi C} [[ \int_0^{\pi}|\sin t|\mathrm{d}t + \int_{\pi}^{2\pi}|\sin t|\mathrm{d}t + \int_{2\pi}^{3\pi}|\sin t|\mathrm{d}t +\cdots \int_{\pi (C-1)}^{\pi (C)}|\sin t|\mathrm{d}t ]] + \frac{1}{A+\pi C}\int_0^{A}|\sin t|\mathrm{d}t$

Pictorially :
|SIN|

Here $x$ is given in terms of $A$ & $C$.

Each Purple Area is the Positive Part of a Cycle of the $\sin$ Curve having $Area=2$.
Each Green Area is the Negative Part of a Cycle of the $\sin$ Curve having $Area=2$.
The last Gray Area is the Partial Cycle having $Area$ between $0$ & $2$.

$D(A,C) = \frac{1}{A+\pi C} [[ 2 + 2 + 2 +\cdots 2 ]] + \frac{1}{A+\pi C}\int_0^{A}|\sin t|\mathrm{d}t$

$D(A,C) = \frac{1}{A+\pi C} [[ 2C ]] + \frac{1}{A+\pi c}\int_0^{A}|\sin t|\mathrm{d}t$

$D(A,C) = \frac{2C}{A+\pi C} + \frac{k}{A+\pi C}$ where $k$ is a number (Depending on $A$) between $0$ & $2$

$\large{\displaystyle\lim_{x\to+\infty}\frac{1}{x}\int_0^x|\sin t|\mathrm{d}t = \lim_{C\to+\infty}D(A,C) = \frac{2}{\pi}}$

The Limit will not change with $A$ or $k$ because the Grey Area is too negligible when $C$ is very large.

Prem
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    Thank you Prem for your detailed answer, I like your answer very much! – Franz Lee Feb 03 '23 at 13:07
  • We might state that the limit is ratio of "Area of full Cycles" to "length of full Cycles" which is ratio of "Area of 1 Cycle" to "length of 1 Cycle" , @FranzLee , Nice to know that it is useful ! – Prem Feb 06 '23 at 05:46